Chapter 2 Solutions
Answers to Questions
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Q14
Q15
Q16
Q17
Q18
Q19
Q20
Q21
Q22
Q23
a.
b.
c.
a.
Speed is distance divided by time, so it will be measured in marsbars/zots.
Velocity has the same units as speed, so it will also be measured in marsbars/zots.
Acceleration is change in speed divided by time, so the units will be marsbars/zots2.
The unit system of inches and days would give velocity units of inches per day and acceleration of inches per
day squared.
b. This would be a terrible choice of units! The distance part would be huge while the time part would be
miniscule!
Since fingernails grow slowly, a unit such as mm/month may be appropriate.
a. The winner of a race must have the greater average speed, so the plodding tortoise is the leader here.
b. Since the hare has the higher average speed taken over short intervals, he is likely to have the greater
instantaneous speed.
In England "doing 70" likely means driving at 90 km/hour which would be a reasonable highway speed. In the US it
means 70 mi/hr. Stating a number without the proper units leaves us uncertain as to what it means.
A speedometer measures instantaneous speed; the speed that you are driving at a particular instant of time. You
can note how it responds immediately as you speed up (accelerate) or slow down (brake).
In low density traffic, the speed is more likely to be constant, therefore the average and the instantaneous speed
will be close for relatively long periods. In high density traffic, the speed is likely to be constantly changing so that
only for short periods the instantaneous speed will equal the average.
The radar gun measures instantaneous speed - the speed at the instant the radar beam hits the car and is reflected
from it. An airplane spotter measures average speed; timing a car between two points which are a known distance
apart.
Yes. When the ball is reflected from the wall (bounces back), the direction of its velocity is different than when it
approaches the wall.
a. Yes. As it moves in a circle the velocity of the ball at each instant is tangent to the circular path. Since this
direction changes, the velocity changes even though the speed remains the same.
b. No. Since the velocity is changing, the acceleration is not zero.
a. No. The velocity changes because the speed and direction of the ball are changing.
b. No. The speed is constantly changing. At the turn-around points the speed is zero.
No. When a ball is dropped it moves in a constant direction (downwards), but the magnitude of its velocity (speed)
increases as it accelerates due to gravity.
Yes. Acceleration is the change in velocity per unit time. Here the change in velocity is negative (the velocity
decreases), so the acceleration is also negative and opposite to the direction of velocity. This is often called a
deceleration.
No. In order to find the acceleration, you need to know the change in velocity that occurs during a time interval.
Knowing the velocity at just one instant tells you nothing about the velocity at a later instant of time.
No. If the car is going to start moving, its acceleration must be non-zero. Otherwise the velocity would not change
and it would remain at rest, or stopped.
No. As the car rounds the curve, the direction of its velocity changes. Since there is a change in velocity (direction
even if not magnitude), it must have an acceleration.
The turtle. As long as the racing car travels with constant velocity (even as large a velocity as 100 MPH), its
acceleration is zero. If the turtle starts to move at all, its velocity will change from zero to something else, and thus
it does have an acceleration.
a. Yes. Constant velocity is represented by the horizontal line which indicates the velocity does not change.
b. Acceleration is greatest between 2 and 4 sec where the slope of the graph is steepest.
a. Yes. The velocity is represented by the slope of a line on a distance-time graph. You can also see that sometime
after pt. B the line has a negative slope indicating a negative velocity.
b. Greater. The instantaneous velocities can be compared by looking at their slopes. The steeper slope indicates
the greater instantaneous velocity.
Yes. The velocity is constant during all three different time intervals, that is in each interval where there is a straight
line. Note that while the velocities are constant in these intervals, they are not the same in each.
a. No. The car has a positive velocity during the entire time shown.
b. At pt. A. The acceleration is greatest since the slope between 0 and 2 sec. is greater than between 4 and 6 sec.
Between 2 and 4 sec. The slope is zero so the velocity in that interval does not change.
Between 2 and 4 sec the car travels the greatest distance. Distance traveled can be determined from a velocitytime graph and is represented by the area under the curve, and between 2 and 4 sec. The area is the largest. The
car travels the next greatest distance between 4 and 6 sec.
a. Yes, during the first part of the motion where the instantaneous speed is greatest. The average speed for the
entire trip must be less than during this interval since for the rest of the trip the speeds are less.
b. Yes. The velocity changes direction. Even if the magnitude of the velocity (speed) is the same, the different
directions make the velocities different. Actually, the change is negative so the car decelerates.
1
Q24
Q25
Q26
Q27
Q28
No. This relationship holds only when the acceleration is constant.
Velocity and distance increase with time when a car accelerates uniformly from rest as long as the acceleration is
positive. As long as it accelerates uniformly, the acceleration is constant.
No. The acceleration is increasing. A constant acceleration is represented by a straight line. Here the curve shown
has an increasing, or positive, slope.
Yes. For uniform acceleration the acceleration is constant. Since acceleration does not change, the average
acceleration equals this constant acceleration.
The distance covered during the first 5 sec is greater than the distance covered during the second 5 sec. Thus
since distance = vot + ½at2, even though acceleration and time are the same for both intervals, the initial velocity at
which the car starts the second interval is less than at the beginning of the first, so it will cover a shorter distance.
Q29
a.
v
t
b.
a
t
Q30
The second runner. If both runners cover the same distance in the same time interval, then their average velocity
has to be the same and the area under the curves on a velocity-time graph are the same. If the first runner reaches
maximum speed quicker, the only way the areas can be equal is if the second runner reaches a higher maximum
speed which he then maintains over a shorter portion of the interval.
Q31
v
second runner
first runner
t
Q32
V
Time
A
Time
2
Answers to Exercises
E1
E2
E3
E4
E5
E6
E7
E8
57.5 MPH
3.6 km/hr
0.4 cm/day
135 mi
200 s
4.84 hr
4.320 km
a. 0.022 km/s
b. 79.2 km/hr
93.3 km/hr
3.5 m/s2
21 m/s
-3 m/s2
a. 17 m/s
b. 29 m
a. 4.4 m/s
b. 6.4 m
a. 21 m/s
b. 76.5 m
a. 3 m/s
b. 3 m
9.09 s
E9
E10
E11
E12
E13
E14
E15
E16
E17
E18
a. Speed: 3 m/s, 6 m/s, 9 m/s, 12 m/s, 15 m/s
b. Distance: 1.5 m, 6 m, 13.5 m, 24 m, 37.5 m
Speed (m/s)
20
15
10
5
0
0
2
4
6
Distance (m)
Tim e (s)
40
35
30
25
20
15
10
5
0
0
1
2
3
4
5
6
Time (s)
3
Answers to Synthesis Problems
SP1
a.
b.
21s
A
v. 5
S 4
p
e
e
d
t (s)
16
c.
v
5
t (s)
-4
d.
a
t (s)
SP2
a.
b.
c.
d.
SP3
a.
1.0 m/s2
2.0 m/s2
1.5 m/s2
No. This acceleration should be calculated as a time average, aav = (a1.t1 + a2.t2)/(t1 + t2) not
aav = (a1 + a2)/2. Note the acceleration here can also be calculated as aav = (v2 - v1)/(t2 - t1).
30
Speed (m/s)
25
20
15
10
5
0
0
20
30
40
Time (s)
4
Acceleration (m/s2)
b.
10
2
0
-2
0
10
20
30
40
-4
4
-6
Tim e (s)
SP4
c.
Yes. The car never has a negative velocity (that is, it moves backwards), so its distance must increase.
a.
b.
c.
5 s since t = (v2 - v1) / a
95 m
Distance (m)
100
80
60
40
20
0
0
1
2
3
4
5
6
Time (s)
Note the parabolic shape of this curve is not obvious over the given time span, but it is indeed a parabola!
SP5
a.
Time (s)
1
2
3
4
5
b.
c.
Distance (m)
Car A
Car B
2.25
9.0
0.25
36
56.2
10
20
30
40
50
Car A passes car B at approximately 4.5 sec.
To find a better time you could graph the distance versus time for each car and see where the two curves
cross. To find the exact time when the distance is the same for both cars, note that then dA = dB = ½ aAt2 = vBt.
Thus tmeet = 0 s and (2 vB/aA ) = 5.0 s.
5